Question: Let $h(x)=3x^4-5x^3+2x^2$. Find $h'(-1)$. Choose 1 answer: Choose 1 answer: (Choice A) A $9$ (Choice B) B $-10$ (Choice C) C $-31$ (Choice D) D $-9$
Explanation: Let's first find the expression for $h'(x)$ and then evaluate it at $x=-1$. According to the sum rule, the derivative of $3x^4-5x^3+2x^2$ is the sum of the derivatives of $3x^4$, $-5x^3$, and $2x^2$. The derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ For example, this is the derivative of the first term: $\begin{aligned}\dfrac{d}{dx}(3x^4)&=3\dfrac{d}{dx}(x^4)&&\gray{\text{Constant multiple rule}}\\\\ &=3\cdot (4x^3)&&\gray{\text{Power rule}}\\ \\ &=12x^3\end{aligned}$ Here is the complete differentiation process: $\begin{aligned} &\phantom{=}h'(x) \\\\ &=\dfrac{d}{dx}(3x^4-5x^3+2x^2) \\\\ &=3\dfrac{d}{dx}(x^4)-5\dfrac{d}{dx}(x^3)+2\dfrac{d}{dx}(x^2)&&\gray{\text{Basic differentiation rules}} \\\\ &=3\cdot 4x^3-5\cdot3x^2+2\cdot 2x&&\gray{\text{The power rule}} \\\\ &=12x^3-15x^2+4x \end{aligned}$ So we found that $h'(x)=12x^3-15x^2+4x$. Plugging in $x=-1$ and evaluating using the calculator, we find that $h'(-1)=-31$. In conclusion, $h'(-1)=-31$.